matrices - Let $A: Bbb{R}^{+}to M_{ntimes n}(Bbb{R})$, then $A(t)$ is continuous if and only if $a_{ij}(t)$ is continuous, $forall;i,j=1,cdots,n$

Suppose \begin{align}A: \Bbb{R}^{+}\to M_{n\times n}(\Bbb{R})\end{align}
\begin{align}t\mapsto A(t),\end{align}
where $A(t)=\big(a_{ij}(t)\big).$

I want to prove that the following are equivalent.

$i.$ $A(t)$ is continuous;

$ii.$ $a_{ij}(t)$ is continuous, $\forall\;i,j=1,\cdots,n$.

MY TRIAL

Assume that $a_{ij}(t)$ is continuous $\forall\;i,j=1,\cdots,n$. Since $M_{n\times n}(\Bbb{R})$ is a finite dimensional vector space, then all norms are equivalent. Take

\begin{align}\Vert A(t)\Vert=\sum^{n}_{i,j=1}|a_{ij}(t)|\end{align}
Since $A(t)$ is a finite sum of continuous functions, then it is continuous.

Now, I need to prove that converse but no way. Please, can anyone help me? Kindly help me check if my proof is as well, correct!

Answer

The last step of your proof for $(ii)\Rightarrow (i)$ does not sound quite right to me. In your last equality, it is $\|A(t)\|$ which is a sum of continuous functions, and hence continuous. In general, this does not imply that $A(t)$ is also continuous.

Hint: consider this equality instead
$$\|A(t)-A(t_0)\|=\sum_{i,j=1}^{n}|a_{ij}(t)-a_{ij}(t_0)| ,\qquad \forall t,t_0\in \mathbb{R}^+$$

and take the limit for $t\to t_0$ (starting from the side of the equality where the limit actually exists).